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I know that $$x\left(\sum\limits_{n\leq x} \frac{\ln(n)}{n^2}\right)+\theta(x) \leq x\left(\sum\limits_{n\leq x} \frac{\ln(n)}{n}\right)+\theta(x)=x\frac{1}{2}\left(\ln{x}\right)^2+Cx+O(\ln x)+O(x)$$

where $\theta(x)=\sum\limits_{p\leq x} \ln p$

Is that equal to $O(x(\ln x)^2)$ not $O(x)$?

rtybase
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Jhon Doe
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1 Answers1

1

Linking this question purely for the links to the theoretical material (and to some extent, similarity).

From Euler's summation formula

$$\sum\limits_{y< n\leq x}f(n)= \int\limits_{y}^{x}f(t)dt + \int\limits_{y}^{x}\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y) \tag{1}$$

and

$$\int \frac{\ln{x}}{x^2} dx=-\frac{\ln{x} + 1}{x} + C \tag{2}$$

we have ($n=1$ is trivial, we will consider $n\geq2$) $$\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}= \int\limits_{2}^{x}\frac{\ln{t}}{t^2}dt+\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt-\{x\}\frac{\ln{x}}{x^2}=\\ \frac{\ln{2}+1}{2}-\frac{\ln{x}+1}{x}+\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt-\{x\}\frac{\ln{x}}{x^2} \tag{3}=...$$

Now, $\forall x\geq2$ $$\left|\{x\}\frac{\ln{x}}{x^2}\right|\leq \frac{\ln{x}}{x^2}$$ and (it's worth noting that $2\ln{2}-1>0$) $$\left|\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt\right| \leq \int\limits_{2}^{x}\left|\{t\}\frac{1-2\ln{t}}{t^3}\right|dt \leq \int\limits_{2}^{x}\left|\frac{1-2\ln{t}}{t^3}\right|dt=\\ \int\limits_{2}^{x}\frac{2\ln{t}-1}{t^3}dt=-\frac{\ln{x}}{x^2}+\frac{\ln{2}}{2^2}$$ this shows that $\int\limits_{2}^{\infty}\{t\}\frac{1-2\ln{t}}{t^3}dt=I$ exists (by taking $\lim\limits_{x\rightarrow\infty}$) and $$\int\limits_{2}^{x}\{t\}\frac{1-2\ln{t}}{t^3}dt=I-\int\limits_{x}^{\infty}\{t\}\frac{1-2\ln{t}}{t^3}dt= I+\int\limits_{x}^{\infty}\{t\}\frac{2\ln{t}-1}{t^3}dt\leq \\ I+\int\limits_{x}^{\infty}\frac{2\ln{t}-1}{t^3}dt=I+\frac{\ln{x}}{x^2}$$

Continuing from $(3)$ $$...=\frac{\ln{2}+1}{2}+I-\frac{\ln{x}+1}{x}+O\left(\frac{\ln{x}}{x^2}\right)= E-\frac{\ln{x}+1}{x}+O\left(\frac{\ln{x}}{x^2}\right) \tag{3a}$$ where $E$ is a constant. From $(3a)$ we see that $$\lim\limits_{x\rightarrow\infty} \left(\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}\right)=E \tag{4}$$ which simply means $$\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}=O(1) \tag{5}$$

Also, Chebyshev's function has the property that $$\lim\limits_{x\rightarrow\infty}\frac{\theta(x)}{x}=1 \Rightarrow \theta(x)=O(x) \tag{6}$$

All these together $$x\left(\sum\limits_{2\leq n\leq x}\frac{\ln{n}}{n^2}\right)+\theta(x)=xO(1)+O(x)=O(x)$$

rtybase
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  • cant you just use integral test to prove the convergence and then proceed from there? – Jhon Doe Feb 10 '19 at 13:55
  • Euler's summation formula, in a way, is a stricter form of integration test. Apostol's "Introduction to Analytic Number Theory" has an entire section dedicated to it (page ~54) – rtybase Feb 10 '19 at 13:57
  • Thanks. I'm new to analytic number theory and i did consider using the summation formula. But the integral test seemed easier. but just to clarify its not wrong to use the integral test it this case to show that the partial sums are bounded right? – Jhon Doe Feb 10 '19 at 14:01
  • You could definitely try it, bounding the summation index to $\left \lfloor x \right \rfloor$, then studying monotonicity of $\frac{\ln{x}}{x^2}$ on $[2,\infty)$, then concluding that $\sum \leq \int ... = const$. It's still a bit of writing :) – rtybase Feb 10 '19 at 14:12